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This salt contains water of crystallization as indicated by the formula, CoCl2.6H2O. On heating, the water of crystallization is lost. The salt first dissolves in this water:
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CoCl2.6H2O(s) ⇒ CoCl2(aq) |
The blue colour of the solution which is formed is due to the formation of a complex ion. Don't worry about this - we shall meet this concept later in the course. Later the water can be seen condensing as colourless drops of liquid at the mouth of the test tube in the video.
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CoCl2(aq) ⇒ CoCl2(s) + (aq) |
The hydrated salt is pink and the anhydrous salt is blue so a colour change from pink to blue is seen on heating the hydrated salt. If water is added to the anhydrous salt, the reverse colour change from blue to pink is seen. This is often made use of as a test for water. Some filter paper is soaked in cobalt(II) chloride solution and heated gently to drive off the water. The paper turns blue, and remains this colour if kept dry. If water is added to the paper a colour change from blue to pink is seen. Other colourless liquids, like ethanol, do not give this colour change.
5 g of CoCl2.6H2O is 5 ÷ 238 = 0.0210 mol
This produces 0.0210 mol of CoCl2. This is 0.0210 × 130 = 2.73 g
This is also a hydrated salt which first loses its water of crystallization. This could be seen condensing and running down near the mouth of the test tube. Most of the metal nitrates decompose on heating to form a metal oxide, nitrogen dioxide and oxygen gases. We detected the nitrogen dioxide gas by its brown colour. This was confirmed when it turned damp, universal indicator red, showing it to be acidic. The oxygen gas produced relit the glowing splint. Copper(II) oxide was the remaining black solid. The equations for this reaction are:
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Cu(NO3)2.6H2O(s) ⇒ Cu(NO3)2(aq) |
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Cu(NO3)2(aq) ⇒ Cu(NO3)2(s) + (aq) |
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2Cu(NO3)2(s) ⇒ 2CuO(s) + 4NO2(g) + O2(g) |
5 g of Cu(NO3)2.6H2O is 5 ÷ 295.5 = 0.0169 mol
This produces 0.0169 mol of CuO. This is 0.0169 × 79.5 = 1.35 g
Water is first released from the crystals to give the anhydrous salt:
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FeSO4.7H2O(s) ⇒ FeSO4(s) + 7H2O(l) |
On further heating a mixture of sulphur dioxide and sulphur trioxide gases are released. These are both very acidic and turn the damp full range indicator red. The reddish-brown solid which remains suggests an iron(III) compound - it is iron(III) oxide:
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2FeSO4(s) ⇒ Fe2O3(s) + SO2(g) + SO3(g) |
5 g of FeSO4.7H2O is 5 ÷ 278 = 0.0180 mol
This produces 0.0180 ÷ 2 = 0.00899 mol of Fe2O3. This is 0.00899 × 160 = 1.44 g
This compound is anhydrous, so no water loss is seen on heating. However, if you look very carefully you will see a small amount of liquid produced - this is water being released from a slightly damp sample. The lime-water is seen to go cloudy. This shows us that carbon dioxide gas was produced by the reaction. This leaves solid zinc oxide:
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ZnCO3(s) ⇒ ZnO(s) + CO2(g) |
Zinc oxide is an unusual compound as its colour is different when hot to that when cold. Hot zinc oxide is yellow and cold zinc oxide is white. In the video we see the colour change from yellow to white as the compound cools down.
5 g of ZnCO3 is 5 ÷ 125.4 = 0.0399 mol
This produces 0.0399 mol of ZnO. This is 0.0210 × 81.4 = 3.25 g
This compound has no water of crystallization. It is first seen melting to produce the liquid. The group I metal nitrates (with the exception of lithium nitrate) are unusual in that they are much more difficult to decompose than other nitrates. On heating, oxygen gas is detected by its effect on a glowing splint. However, no nitrogen dioxide is produced as was the case with copper(II) nitrate. The pale yellow solid which remains, on cooling, is sodium nitrite (the modern name is sodium nitrate(III)).
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NaNO3(s) ⇒ NaNO3(l) |
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NaNO3(l) ⇒ NaNO2(l) + ½O2(g) |
5 g of NaNO3 is 5 ÷ 85 = 0.0588 mol
This produces 0.0588 mol of NaNO2. This is 0.0588 × 69 = 4.06 g